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MHT CET202124 Sep 2021Evening ShiftMathematicsCircleActual

If y=2 x is a chord of circle x^2+y^2-10 x=0 , then the equation of circle with this chord as diameter is

Options

  1. Ax^2+y^2-2 x-4 y=0
  2. Bx^2+y^2+2 x+4 y=0
  3. Cx^2+y^2-2 x+4 y=0
  4. Dx^2+y^2+2 x-4 y=0

Correct answer

A. x^2+y^2-2 x-4 y=0

Step-by-step solution

aligned & x^2-10 x+y^2=0 & x-10 x+25+y^2=25 centre =(5,0) and r=5 aligned y =2 x is a chord of given circle. Point of intersection of chord and circle is x ^2-10 x +25+4 x ^2=25 y =0,4 Thus end points of the chord are (0,0) and (2,4) Mid point of the chord = ( 2 2 , 4 2 )=(1,2) and length of chord = (2)^2+(4)^2 = 20 is the diameter of required circle. Hence equation of required circle is (x-1)+(y-2)= ( 20 2 )^2 i.e. x^2+y^2-2 x-4 y=0

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